CBSE Class 10 Mathematics Quadratic Equations Assignment Set 02

Important Concepts

Take a Look:

• A quadratic equation in its standard form is written as \( ax^2+bx+c=0 \), where \( a \neq 0 \). Here, \( a \), \( b \), and \( c \) represent real coefficients.
• Any real value \( x \) that satisfies \( ax^2+bx+c=0 \) acts as a root. Furthermore, the zeroes of a quadratic expression are identical to the solutions of its corresponding quadratic equation.
• The quantity \( b^2-4ac \) is defined as the discriminant, commonly represented by the letter \( D \). Therefore, the formula is \( D = b^2-4ac \).
• Two roots exist for any quadratic equation; these roots can be real and distinct, equal, or complex numbers.
• If \( \alpha \) and \( \beta \) are the roots, we find them using the quadratic formula:
  \( \alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \) and \( \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \).
• The sum of these roots is given by \( \alpha + \beta = -\frac{b}{a} \), and their product is \( \alpha\beta = \frac{c}{a} \).
• When given the roots, a quadratic equation is constructed using \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \).
• How the roots behave depends on the discriminant \( D \):
  • When \( D > 0 \), the equation has two distinct real roots.
  • When \( D = 0 \), the roots are real and identical.
  • When \( D < 0 \), there are no real roots.
     

Level-I
 

Question. IF ½ is a root of the equation x2+kx-5/4=0, then the value of K is
(a) 2
(b) -2
(c) ¼
(d) ½
Answer: (a) 2
In simple words: Substituting the root \( \frac{1}{2} \) into the equation allows us to solve for \( k \), which gives \( 2 \).

Exam Tip: Since substituting a root makes the quadratic equation equal to zero, simply replace \( x \) with the given root value and solve for the unknown coefficient.

Question. IF D>0, then roots of a quadratic equation ax2+bx+c=0 are
(a) \( \frac{-b \pm \sqrt{D}}{2a} \)
(b) \( \frac{b + \sqrt{D}}{2a} \)
(c) \( \frac{b - \sqrt{D}}{2a} \)
(d) None of the options
Answer: (a) \( \frac{-b \pm \sqrt{D}}{2a} \)
In simple words: When the discriminant is positive, the quadratic formula provides two distinct real roots.

Exam Tip: Remember that the plus-minus sign in the quadratic formula accounts for both roots of the equation when the discriminant is positive.

Question. Discriminant of x2 +5x+5=0 is
(a) 5/2
(b) -5
(c) 5
(d) -4
Answer: (c) 5
In simple words: Calculating \( b^2 - 4ac \) with \( a=1 \), \( b=5 \), and \( c=5 \) gives \( 25 - 20 = 5 \).

Exam Tip: Keep the signs of \( a \), \( b \), and \( c \) intact when inserting them into the formula \( D = b^2 - 4ac \).

Question. The sum of roots of a quadratic equation x2 +4x-320=0 is
(a) -4
(b) 4
(c) 1/4
(d) 1/2
Answer: (a) -4
In simple words: Using the formula \( -\frac{b}{a} \), we divide \( -4 \) by \( 1 \) to get a sum of \( -4 \).

Exam Tip: Be careful with the negative sign in the sum of roots formula \( -\frac{b}{a} \); a positive \( b \) value results in a negative sum when \( a \) is positive.

Question. The product of roots of a quaradatic equation 2 x2 +7x-4=0 is
(a) 2/7
(b) -2/7
(c) -4/7
(d) -2
Answer: (d) -2
In simple words: The formula for the product of roots is \( \frac{c}{a} \), which gives \( \frac{-4}{2} = -2 \).

Exam Tip: The product of roots is always the constant term divided by the coefficient of \( x^2 \), keeping their original signs.

Question. Values of K for which the equation 9 x2 +2kx-1=0 has real roots are:
(a) \( k \ge \pm 3 \)
(b) \( k \ge 3 \) or \( K \le -3 \)
(c) \( K \ge -3 \)
(d) \( k \le \pm 3 \)
Answer: (b) \( k \ge 3 \) or \( K \le -3 \)
In simple words: For the equation to have real roots, the discriminant \( D = (2k)^2 - 4(9)(1) \) must be greater than or equal to zero, which gives \( k \ge 3 \) or \( k \le -3 \).

Exam Tip: For inequalities of the form \( x^2 \ge a^2 \), the solution is always \( x \ge a \) or \( x \le -a \).

Level-II
 

Question. Find the roots of the quadratic equation 3 x2 -2√6x+2=0
Answer:
Given equation: \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
Splitting the middle term, we get:
\( 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \)
\( \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \)
\( (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0 \)
This gives:
\( \sqrt{3}x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} = \frac{2\sqrt{6}}{6} \)
Hence, the roots are \( x = \frac{2\sqrt{6}}{6}, \frac{2\sqrt{6}}{6} \).
In simple words: We factor the equation by splitting the middle term to get two identical roots, which simplify to \( \frac{2\sqrt{6}}{6} \).

Exam Tip: For equations involving square roots, factor the coefficients using their square root components to easily group and solve them.

Question. The sum of the squares of two consecutive odd number is 394. Find the numbers.
Answer:
Let the consecutive odd integers be \( x \) and \( x + 2 \).
According to the problem:
\( x^2 + (x + 2)^2 = 394 \)
\( x^2 + x^2 + 4x + 4 = 394 \)
\( 2x^2 + 4x + 4 - 394 = 0 \)
\( 2x^2 + 4x - 390 = 0 \)
Dividing the entire equation by \( 2 \):
\( x^2 + 2x - 195 = 0 \)
Factoring the quadratic equation:
\( x^2 + 15x - 13x - 195 = 0 \)
\( x(x + 15) - 13(x + 15) = 0 \)
\( (x - 13)(x + 15) = 0 \)
So, \( x = 13 \) or \( x = -15 \).
If \( x = 13 \), the consecutive odd numbers are \( 13 \) and \( 15 \).
If \( x = -15 \), the consecutive odd numbers are \( -15 \) and \( -13 \).
Therefore, the required numbers are \( 13, 15 \) or \( -15, -13 \).
In simple words: We set up the sum of squares for two consecutive odd numbers as a quadratic equation, which gives the solutions \( 13, 15 \) or \( -15, -13 \).

Exam Tip: Don't forget that odd numbers can be negative too, so you must list both positive and negative pairs of solutions.

Question. Find the root of 6 x2 +√2x-2=0 by factorization.
Answer:
Given equation: \( 6x^2 + \sqrt{2}x - 2 = 0 \)
We need to find two numbers whose product is \( 6 \times (-2) = -12 \) and whose sum is \( \sqrt{2} \). These numbers are \( 3\sqrt{2} \) and \( -2\sqrt{2} \).
Splitting the middle term:
\( 6x^2 + 3\sqrt{2}x - 2\sqrt{2}x - 2 = 0 \)
Taking out common factors:
\( 3\sqrt{2}x(\sqrt{2}x + 1) - 2(\sqrt{2}x + 1) = 0 \)
So:
\( (3\sqrt{2}x - 2)(\sqrt{2}x + 1) = 0 \)
Setting each factor to zero:
\( 3\sqrt{2}x - 2 = 0 \implies x = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} \)
\( \sqrt{2}x + 1 = 0 \implies x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \)
Thus, the roots are \( x = \frac{\sqrt{2}}{3} \) and \( x = -\frac{\sqrt{2}}{2} \).
In simple words: We split the middle term into \( 3\sqrt{2}x \) and \( -2\sqrt{2}x \), group the terms, and factor them to find the roots \( \frac{\sqrt{2}}{3} \) and \( -\frac{\sqrt{2}}{2} \).

Exam Tip: Rationalizing the denominator of your final roots is a good habit that ensures your answers match standard marking schemes.

Question. The sum of two numbers is 8 . Determine the numbers if the sum of their reciprocals is 8/15.
Answer:
Let the two numbers be \( x \) and \( 8 - x \).
According to the given condition:
\( \frac{1}{x} + \frac{1}{8-x} = \frac{8}{15} \)
Taking the LCM on the left side:
\( \frac{(8-x) + x}{x(8-x)} = \frac{8}{15} \)
\( \frac{8}{8x - x^2} = \frac{8}{15} \)
Dividing both sides by \( 8 \):
\( \frac{1}{8x - x^2} = \frac{1}{15} \)
Cross-multiplying:
\( 8x - x^2 = 15 \)
\( x^2 - 8x + 15 = 0 \)
Factoring the quadratic equation:
\( x^2 - 5x - 3x + 15 = 0 \)
\( x(x - 5) - 3(x - 5) = 0 \)
\( (x - 3)(x - 5) = 0 \)
This gives \( x = 3 \) or \( x = 5 \).
If \( x = 3 \), the numbers are \( 3 \) and \( 5 \).
If \( x = 5 \), the numbers are \( 5 \) and \( 3 \).
Therefore, the two numbers are \( 3 \) and \( 5 \).
In simple words: Representing the two numbers as \( x \) and \( 8-x \) allows us to set up their reciprocals in an equation that reduces to \( x^2 - 8x + 15 = 0 \), giving the numbers \( 3 \) and \( 5 \).

Exam Tip: Simplify the algebraic fraction by cross-multiplying early on to avoid dealing with complex denominators.

Question. For what value of k does (k-12)x2+2(k-12)x+2=0 has equal roots?
Answer:
Given equation: \( (k-12)x^2 + 2(k-12)x + 2 = 0 \)
For a quadratic equation to have equal roots, its discriminant \( D \) must be \( 0 \).
Here, \( a = k-12 \), \( b = 2(k-12) \), and \( c = 2 \).
\( D = b^2 - 4ac = 0 \)
\( [2(k-12)]^2 - 4(k-12)(2) = 0 \)
\( 4(k-12)^2 - 8(k-12) = 0 \)
Dividing by \( 4 \):
\( (k-12)^2 - 2(k-12) = 0 \)
Taking \( (k-12) \) as a common factor:
\( (k-12)[(k-12) - 2] = 0 \)
\( (k-12)(k-14) = 0 \)
This gives \( k = 12 \) or \( k = 14 \).
However, if \( k = 12 \), then the coefficient of \( x^2 \) becomes \( 0 \nolinebreak \), which means the equation is no longer quadratic.
Therefore, we reject \( k = 12 \).
Thus, the required value of \( k \) is \( 14 \).
In simple words: Setting the discriminant to zero yields \( k=12 \) and \( k=14 \). We rule out \( k=12 \) because it would eliminate the \( x^2 \) term, leaving \( k=14 \) as our answer.

Exam Tip: Always check if your solved value of the variable makes the leading coefficient zero; if it does, that value must be discarded because a quadratic equation must have \( a \neq 0 \).

Question. Divide 51 into two parts whose product is 378.
Answer:
Let one part of \( 51 \) be \( x \). The other part will be \( 51 - x \).
According to the problem:
\( x(51 - x) = 378 \)
\( 51x - x^2 = 378 \)
\( x^2 - 51x + 378 = 0 \)
Factoring the quadratic equation:
\( x^2 - 42x - 9x + 378 = 0 \)
\( x(x - 42) - 9(x - 42) = 0 \)
\( (x - 9)(x - 42) = 0 \)
This gives \( x = 9 \) or \( x = 42 \).
Thus, the two parts are \( 9 \) and \( 42 \).
In simple words: Splitting \( 51 \) into \( x \) and \( 51-x \) gives us a quadratic equation with solutions \( 9 \) and \( 42 \), which are the two parts.

Exam Tip: When splitting a number into two parts, write them as \( x \) and \( (\text{total} - x) \) to keep the equation in a single variable.

Level-III
 

Question. For what value of k, will the equation 2 x2 -2(1+2k)x+(3+2k)=0 have real but distinct roots ? When will the roots be equal?
Answer:
Given quadratic equation: \( 2x^2 - 2(1+2k)x + (3+2k) = 0 \)
Here, \( a = 2 \), \( b = -2(1+2k) \), and \( c = 3+2k \).
The discriminant \( D \) is calculated as:
\( D = b^2 - 4ac \)
\( D = [-2(1+2k)]^2 - 4(2)(3+2k) \)
\( D = 4(1 + 4k + 4k^2) - 8(3 + 2k) \)
\( D = 4 + 16k + 16k^2 - 24 - 16k \)
\( D = 16k^2 - 20 \)
For the roots to be real and distinct, we must have \( D > 0 \):
\( 16k^2 - 20 > 0 \)
\( 16k^2 > 20 \)
\( k^2 > \frac{20}{16} \)
\( k^2 > \frac{5}{4} \)
Taking the square root on both sides:
\( k > \frac{\sqrt{5}}{2} \) or \( k < -\frac{\sqrt{5}}{2} \).

For the roots to be equal, we must have \( D = 0 \):
\( 16k^2 - 20 = 0 \)
\( k^2 = \frac{5}{4} \)
\( k = \pm\frac{\sqrt{5}}{2} \).
Therefore, the roots are real and distinct when \( k > \frac{\sqrt{5}}{2} \) or \( k < -\frac{\sqrt{5}}{2} \), and they are equal when \( k = \pm\frac{\sqrt{5}}{2} \).
In simple words: Setting the discriminant \( D > 0 \) gives \( k > \frac{\sqrt{5}}{2} \) or \( k < -\frac{\sqrt{5}}{2} \) for distinct roots, while \( D = 0 \) gives \( k = \pm\frac{\sqrt{5}}{2} \) for equal roots.

Exam Tip: For inequalities of the form \( x^2 > a^2 \), the solution is always \( x > a \) or \( x < -a \).

Question. Solve for x: 4√3x2 +5x-2√3=0.
Answer:
Given equation: \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
We split the middle term using the numbers \( 8 \) and \( -3 \), which multiply to \( 4\sqrt{3} \times (-2\sqrt{3}) = -24 \) and add up to \( 5 \):
\( 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \)
Group the terms:
\( 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \)
\( (4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \)
Setting each factor to zero:
\( 4x - \sqrt{3} = 0 \implies x = \frac{\sqrt{3}}{4} \)
\( \sqrt{3}x + 2 = 0 \implies x = -\frac{2}{\sqrt{3}} \)
Hence, the solutions are \( x = \frac{\sqrt{3}}{4} \) and \( x = -\frac{2}{\sqrt{3}} \).
In simple words: By splitting the middle term into \( 8x \) and \( -3x \), we factor the expression to get the solutions \( \frac{\sqrt{3}}{4} \) and \( -\frac{2}{\sqrt{3}} \).

Exam Tip: When factoring quadratic equations with roots, look for terms that can combine to form the integer coefficients of the equation.

Question. Using quadratic formula solve the following quadratic equation for x: x2-2ax+(a2-b2)=0
Answer:
Given equation: \( x^2 - 2ax + (a^2 - b^2) = 0 \)
Compare this with the standard quadratic equation \( Ax^2 + Bx + C = 0 \):
\( A = 1 \), \( B = -2a \), and \( C = a^2 - b^2 \).
First, find the discriminant \( D \):
\( D = B^2 - 4AC \)
\( D = (-2a)^2 - 4(1)(a^2 - b^2) \)
\( D = 4a^2 - 4a^2 + 4b^2 \)
\( D = 4b^2 \)
Using the quadratic formula:
\( x = \frac{-B \pm \sqrt{D}}{2A} \)
\( x = \frac{2a \pm \sqrt{4b^2}}{2(1)} \)
\( x = \frac{2a \pm 2b}{2} \)
\( x = a \pm b \)
Thus, the roots of the equation are \( x = a + b \) and \( x = a - b \).
In simple words: Substituting the coefficients into the quadratic formula simplifies the discriminant to \( 4b^2 \), yielding the roots \( a + b \) and \( a - b \).

Exam Tip: Do not expand \( (a^2 - b^2) \) unnecessarily; keeping it grouped helps simplify the discriminant quickly.

Question. The speed of a boat in still water is 11 km/hr . It can go 12 km up stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.
Answer:
Let the speed of the stream be \( y \) km/hr.
Given:
Speed of the boat in still water = \( 11 \) km/hr.
So, speed of the boat upstream = \( (11 - y) \) km/hr.
Speed of the boat downstream = \( (11 + y) \) km/hr.
The total distance traveled each way is \( 12 \) km.
The total time taken is \( 2 \text{ hours } 45 \text{ minutes} = 2\frac{45}{60} \text{ hours} = 2\frac{3}{4} \text{ hours} = \frac{11}{4} \text{ hours} \).
According to the given condition:
\( \frac{12}{11 - y} + \frac{12}{11 + y} = \frac{11}{4} \)
\( 12 \left[ \frac{(11 + y) + (11 - y)}{(11 - y)(11 + y)} \right] = \frac{11}{4} \)
\( 12 \left[ \frac{22}{121 - y^2} \right] = \frac{11}{4} \)
\( \frac{12 \times 2}{121 - y^2} = \frac{1}{4} \)
\( \frac{24}{121 - y^2} = \frac{1}{4} \)
Cross-multiplying:
\( 121 - y^2 = 96 \)
\( y^2 = 121 - 96 \)
\( y^2 = 25 \)
\( y = 5 \) (since speed of the stream cannot be negative).
Therefore, the speed of the stream is \( 5 \) km/hr.
In simple words: We set up a time equation using the upstream and downstream speeds, which yields a quadratic equation with a positive solution of \( 5 \) km/hr.

Exam Tip: Convert time into a single unit (hours) before writing down your speed-time equation.

Question. Solve for x: a2b2x2+b2x-a2x-1=0
Answer:
Given equation: \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \)
Factor the equation by grouping:
\( b^2x(a^2x + 1) - 1(a^2x + 1) = 0 \)
\( (b^2x - 1)(a^2x + 1) = 0 \)
Setting each factor to zero:
\( b^2x - 1 = 0 \implies x = \frac{1}{b^2} \)
\( a^2x + 1 = 0 \implies x = -\frac{1}{a^2} \)
Thus, the solutions are \( x = \frac{1}{b^2} \) and \( x = -\frac{1}{a^2} \).
In simple words: Grouping terms in pairs allows us to factor out \( (a^2x + 1) \), leading to the root values of \( \frac{1}{b^2} \) and \( -\frac{1}{a^2} \).

Exam Tip: Grouping is highly effective for algebraic quadratics; look for common variables in the first two and last two terms.

Question. Solve the following quadratic equation for x: X2-2(a+2)x+(a+1)(a+3)=0
Answer:
Given equation: \( x^2 - 2(a+2)x + (a+1)(a+3) = 0 \)
Notice that the sum of the constant factors is:
\( (a + 1) + (a + 3) = 2a + 4 = 2(a + 2) \)
Splitting the middle term accordingly:
\( x^2 - [(a + 1) + (a + 3)]x + (a + 1)(a + 3) = 0 \)
\( x^2 - (a + 1)x - (a + 3)x + (a + 1)(a + 3) = 0 \)
Grouping the terms:
\( x[x - (a + 1)] - (a + 3)[x - (a + 1)] = 0 \)
\( [x - (a + 1)][x - (a + 3)] = 0 \)
Setting each factor to zero:
\( x - (a + 1) = 0 \implies x = a + 1 \)
\( x - (a + 3) = 0 \implies x = a + 3 \)
Hence, the roots of the equation are \( x = a + 1 \) and \( x = a + 3 \).
In simple words: Since the sum of the factors \( (a+1) \) and \( (a+3) \) equals the middle term coefficient, we factor the equation directly to find the roots \( a+1 \) and \( a+3 \).

Exam Tip: Look for a direct relationship between the constant term's factors and the middle coefficient before using the quadratic formula.

Level-IV
 

Question. Solve for x: \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \); \( a \neq 0, b \neq 0, x \neq 0 \)
Answer:
Given equation:
\( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
Rearranging the terms by bringing \( \frac{1}{x} \) to the left side:
\( \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
Taking the LCM on both sides:
\( \frac{x - (a + b + x)}{x(a + b + x)} = \frac{b + a}{ab} \)
\( \frac{x - a - b - x}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \)
Dividing both sides by \( (a + b) \) (assuming \( a + b \neq 0 \)):
\( \frac{-1}{x^2 + ax + bx} = \frac{1}{ab} \)
Cross-multiplying:
\( x^2 + ax + bx = -ab \)
\( x^2 + ax + bx + ab = 0 \)
Factoring by grouping:
\( x(x + a) + b(x + a) = 0 \)
\( (x + a)(x + b) = 0 \)
This gives \( x = -a \) or \( x = -b \).
Hence, the solutions are \( x = -a \) and \( x = -b \).
In simple words: Moving \( \frac{1}{x} \) to the left side and taking common denominators simplifies the expression into \( (x+a)(x+b) = 0 \), which gives the roots \( -a \) and \( -b \).

Exam Tip: Grouping the variable \( x \) on one side is the crucial first step to simplify this type of rational quadratic equation.

Question. An aeroplane left 30 minutes later than its scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Determine its usual speed.
Answer:
Let the standard speed of the aeroplane be \( x \) km/hr.
Given:
Distance to destination = \( 1500 \) km.
Increased speed = \( (x + 250) \) km/hr.
The difference in travel times is \( 30 \) minutes, which is \( \frac{1}{2} \) hour.
According to the given condition:
\( \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \)
\( 1500 \left[ \frac{(x + 250) - x}{x(x + 250)} \right] = \frac{1}{2} \)
\( \frac{1500 \times 250}{x^2 + 250x} = \frac{1}{2} \)
\( \frac{375000}{x^2 + 250x} = \frac{1}{2} \)
Cross-multiplying:
\( x^2 + 250x = 750000 \)
\( x^2 + 250x - 750000 = 0 \)
Factoring the equation:
\( x^2 + 1000x - 750x - 750000 = 0 \)
\( x(x + 1000) - 750(x + 1000) = 0 \)
\( (x - 750)(x + 1000) = 0 \)
So, \( x = 750 \) or \( x = -1000 \).
Since speed cannot be negative, we reject \( x = -1000 \).
Therefore, the usual speed of the aeroplane is \( 750 \) km/hr.
In simple words: Setting up a time equation with the distance of \( 1500 \) km and speed difference of \( 250 \) km/hr yields a quadratic equation with a realistic positive speed of \( 750 \) km/hr.

Exam Tip: Word problems regarding speed always have only one positive, physically possible solution; reject any negative results with a brief note.

Question. Using the quadratic formula solve the equation a2b2x2–(4b4-3a4)x -12a2b2=0
Answer:
Given equation: \( a^2b^2x^2 - (4b^4 - 3a^4)x - 12a^2b^2 = 0 \)
Comparing this with \( Ax^2 + Bx + C = 0 \):
\( A = a^2b^2 \)
\( B = -(4b^4 - 3a^4) \)
\( C = -12a^2b^2 \)
Now, find the discriminant \( D \):
\( D = B^2 - 4AC \)
\( D = [-(4b^4 - 3a^4)]^2 - 4(a^2b^2)(-12a^2b^2) \)
\( D = (4b^4 - 3a^4)^2 + 48a^4b^4 \)
Expanding the squared term:
\( D = 16b^8 - 24a^4b^4 + 9a^8 + 48a^4b^4 \)
\( D = 16b^8 + 24a^4b^4 + 9a^8 \)
This is a perfect square:
\( D = (4b^4 + 3a^4)^2 \)
So, \( \sqrt{D} = 4b^4 + 3a^4 \).
Using the quadratic formula:
\( x = \frac{-B \pm \sqrt{D}}{2A} \)
\( x = \frac{(4b^4 - 3a^4) \pm (4b^4 + 3a^4)}{2a^2b^2} \)

Case 1: Using the positive sign:
\( x = \frac{4b^4 - 3a^4 + 4b^4 + 3a^4}{2a^2b^2} = \frac{8b^4}{2a^2b^2} = \frac{4b^2}{a^2} \)

Case 2: Using the negative sign:
\( x = \frac{4b^4 - 3a^4 - (4b^4 + 3a^4)}{2a^2b^2} = \frac{-6a^4}{2a^2b^2} = -\frac{3a^2}{b^2} \)
Therefore, the roots are \( x = -\frac{3a^2}{b^2} \) and \( x = \frac{4b^2}{a^2} \).
In simple words: Applying the quadratic formula simplifies the discriminant to the perfect square \( (4b^4 + 3a^4)^2 \), giving the roots \( -\frac{3a^2}{b^2} \) and \( \frac{4b^2}{a^2} \).

Exam Tip: Recognizing algebraic identities like \( (x-y)^2 + 4xy = (x+y)^2 \) helps to quickly simplify the discriminant under the square root.

Question. Solve for x: \( \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} \) (\( x \neq 2,4 \))
Answer:
Given equation:
\( \frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3} \)
Taking the LCM on the left side:
\( \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3} \)
\( \frac{(x^2 - 5x + 4) + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3} \)
\( \frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3} \)
Dividing the numerator on the left side by \( 2 \):
\( \frac{x^2 - 5x + 5}{x^2 - 6x + 8} = \frac{5}{3} \)
Cross-multiplying:
\( 3(x^2 - 5x + 5) = 5(x^2 - 6x + 8) \)
\( 3x^2 - 15x + 15 = 5x^2 - 30x + 40 \)
Rearranging to form a quadratic equation:
\( 2x^2 - 15x + 25 = 0 \)
Factoring the equation:
\( 2x^2 - 10x - 5x + 25 = 0 \)
\( 2x(x - 5) - 5(x - 5) = 0 \)
\( (2x - 5)(x - 5) = 0 \)
This gives \( x = \frac{5}{2} \) or \( x = 5 \).
Hence, the solutions are \( x = \frac{5}{2} \) and \( x = 5 \).
In simple words: Combining the fractions on the left side and cross-multiplying leads to the quadratic equation \( 2x^2 - 15x + 25 = 0 \), which has roots \( \frac{5}{2} \) and \( 5 \).

Exam Tip: Always state the given restrictions (like \( x \neq 2,4 \)) in your final steps to confirm your solutions are valid.

Question. If (-5) is a root of the quadratic equation 2x2+px-15=0 and the quadratic equation p(x2+x)+k=0 has equal roots then find the values of P and K.
Answer:
Since \( -5 \) is a root of the equation \( 2x^2 + px - 15 = 0 \), it must satisfy it:
\( 2(-5)^2 + p(-5) - 15 = 0 \)
\( 2(25) - 5p - 15 = 0 \)
\( 50 - 5p - 15 = 0 \)
\( 35 - 5p = 0 \)
\( 5p = 35 \implies p = 7 \).

Substituting \( p = 7 \) into the second quadratic equation:
\( 7(x^2 + x) + k = 0 \)
\( 7x^2 + 7x + k = 0 \)
For this equation to have equal roots, its discriminant \( D \) must be \( 0 \):
\( D = b^2 - 4ac = 0 \)
\( 7^2 - 4(7)(k) = 0 \)
\( 49 - 28k = 0 \)
\( 28k = 49 \)
\( k = \frac{49}{28} = \frac{7}{4} \).
Therefore, the values are \( p = 7 \) and \( k = \frac{7}{4} \).
In simple words: Substituting \( -5 \) into the first equation gives \( p = 7 \). Using \( p = 7 \) in the second equation and setting the discriminant to zero yields \( k = \frac{7}{4} \).

Exam Tip: Be methodical; solve for the single unknown variable first, then plug that value into the second equation to find the other variable.

Question. The sum or the areas of two squares is 640m2. If the difference of their perimeters is 64m . Find the sides of the two squares.
Answer:
Let the side of the larger square be \( x \) meters and the side of the smaller square be \( y \) meters.
Given:
Sum of their areas = \( 640 \text{ m}^2 \implies x^2 + y^2 = 640 \)
Difference of their perimeters = \( 64 \text{ m} \implies 4x - 4y = 64 \)
Dividing by \( 4 \):
\( x - y = 16 \implies x = y + 16 \)
Substitute \( x = y + 16 \) into the area equation:
\( (y + 16)^2 + y^2 = 640 \)
\( y^2 + 32y + 256 + y^2 = 640 \)
\( 2y^2 + 32y + 256 - 640 = 0 \)
\( 2y^2 + 32y - 384 = 0 \)
Dividing the equation by \( 2 \):
\( y^2 + 16y - 192 = 0 \)
Factoring the quadratic equation:
\( y^2 + 24y - 8y - 192 = 0 \)
\( y(y + 24) - 8(y + 24) = 0 \)
\( (y - 8)(y + 24) = 0 \)
This gives \( y = 8 \) or \( y = -24 \).
Since the side length of a square cannot be negative, we reject \( y = -24 \).
So, \( y = 8 \text{ m} \).
Substituting \( y = 8 \) into \( x = y + 16 \):
\( x = 8 + 16 = 24 \text{ m} \).
Therefore, the sides of the two squares are \( 24 \text{ m} \) and \( 8 \text{ m} \).
In simple words: Expressing the side of one square in terms of the other lets us create a quadratic equation for the areas, which yields sides of \( 24 \text{ m} \) and \( 8 \text{ m} \).

Exam Tip: Don't forget to include the unit (meters) in your final answers for word problems concerning dimensions.

Self Evaluation
 

Question. If the root of the equation (b-c)x2+(c-a)x+(a-b)=0 are equal, then prove that 2b=a+c
Answer:
Given quadratic equation: \( (b-c)x^2 + (c-a)x + (a-b) = 0 \)
Comparing this with the standard quadratic form \( Ax^2 + Bx + C = 0 \):
\( A = b-c \), \( B = c-a \), and \( C = a-b \).
For the roots to be equal, the discriminant \( D \) must be equal to zero:
\( D = B^2 - 4AC = 0 \)
Substitute the values:
\( (c-a)^2 - 4(b-c)(a-b) = 0 \)
Expanding the terms:
\( (c^2 - 2ca + a^2) - 4(ba - b^2 - ca + cb) = 0 \)
\( a^2 - 2ca + c^2 - 4ab + 4b^2 + 4ca - 4bc = 0 \)
Combine like terms:
\( a^2 + 4b^2 + c^2 - 4ab - 4bc + 2ca = 0 \)
This expression can be rewritten as a perfect square:
\( (a - 2b + c)^2 = 0 \)
Taking the square root on both sides:
\( a - 2b + c = 0 \)
\( 2b = a + c \)
Hence proved.
In simple words: Setting the discriminant to zero leads to the algebraic expansion \( (a-2b+c)^2 = 0 \), which simplifies directly to \( 2b = a+c \).

Exam Tip: Remember the algebraic identity \( (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx \) to factorize trinomial quadratics easily.

Question. Solve by using quadratic formula (x2+3x+2)2- 8(x2+3x)-4=0
Answer:
Given equation: \( (x^2 + 3x + 2)^2 - 8(x^2 + 3x) - 4 = 0 \)
Let \( y = x^2 + 3x \). Substituting this into the equation:
\( (y + 2)^2 - 8y - 4 = 0 \)
\( y^2 + 4y + 4 - 8y - 4 = 0 \)
\( y^2 - 4y = 0 \)
\( y(y - 4) = 0 \)
This gives \( y = 0 \) or \( y = 4 \).

Case 1: When \( y = 0 \):
\( x^2 + 3x = 0 \nolinebreak \)
Using the quadratic formula where \( a = 1 \), \( b = 3 \), and \( c = 0 \):
\( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(0)}}{2(1)} = \frac{-3 \pm 3}{2} \)
This yields \( x = 0 \) or \( x = -3 \).

Case 2: When \( y = 4 \):
\( x^2 + 3x - 4 = 0 \)
Using the quadratic formula where \( a = 1 \), \( b = 3 \), and \( c = -4 \):
\( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \)
This yields \( x = 1 \) or \( x = -4 \).

Therefore, the solutions are \( x = 0, -3, 1, -4 \).
In simple words: Substituting \( y = x^2 + 3x \) simplifies the equation to \( y^2 - 4y = 0 \nolinebreak \), and solving the resulting quadratics for \( x \) gives the values \( 0, -3, 1, -4 \).

Exam Tip: Substituting a recurring quadratic term with a new variable makes complex higher-degree equations much simpler to solve.

Question. If \(\alpha\) and \(\beta\) are the roots of the equation lx2-mx+n=0, Find the equation whose roots are \(\alpha/\beta\) and \(\beta/\alpha\).
Answer:
Given equation: \( lx^2 - mx + n = 0 \)
Since \( \alpha \) and \( \beta \) are the roots of this equation, we have:
Sum of roots: \( \alpha + \beta = \frac{m}{l} \)
Product of roots: \( \alpha\beta = \frac{n}{l} \)

We need to find the quadratic equation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \).
First, find the sum of the new roots:
\( S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Substitute the values:
\( S = \frac{\left(\frac{m}{l}\right)^2 - 2\left(\frac{n}{l}\right)}{\frac{n}{l}} = \frac{\frac{m^2}{l^2} - \frac{2n}{l}}{\frac{n}{l}} = \frac{\frac{m^2 - 2nl}{l^2}}{\frac{n}{l}} = \frac{m^2 - 2nl}{ln} \trim \)

Next, find the product of the new roots:
\( P = \left(\frac{\alpha}{\beta}\right) \times \left(\frac{\beta}{\alpha}\right) = 1 \)

The required quadratic equation is given by:
\( x^2 - Sx + P = 0 \)
\( x^2 - \left(\frac{m^2 - 2nl}{ln}\right)x + 1 = 0 \)
Multiplying the entire equation by \( ln \):
\( ln x^2 - (m^2 - 2nl)x + ln = 0 \)
Therefore, the required equation is \( ln x^2 - (m^2 - 2nl)x + ln = 0 \).
In simple words: We calculate the sum and product of the new roots using the coefficients of the given equation, yielding the final quadratic equation \( ln x^2 - (m^2 - 2nl)x + ln = 0 \).

Exam Tip: Always express the sum of the squares of roots in terms of \( \alpha+\beta \) and \( \alpha\beta \) as \( (\alpha+\beta)^2-2\alpha\beta \) when forming new equations.

Question. The difference of two numbers in 5 and the difference of their reciprocal is 1/10.Find the numbers.
Answer:
Let the two numbers be \( x \) and \( y \), where \( x > y \).
According to the first condition:
\( x - y = 5 \implies x = y + 5 \)
According to the second condition, the difference of their reciprocals is \( \frac{1}{10} \):
\( \frac{1}{y} - \frac{1}{x} = \frac{1}{10} \)
Taking LCM on the left side:
\( \frac{x - y}{xy} = \frac{1}{10} \)
Substitute \( x - y = 5 \):
\( \frac{5}{xy} = \frac{1}{10} \)
Cross-multiplying:
\( xy = 50 \)
Substitute \( x = y + 5 \) into the product equation:
\( (y + 5)y = 50 \)
\( y^2 + 5y - 50 = 0 \)
Factoring the quadratic equation:
\( y^2 + 10y - 5y - 50 = 0 \)
\( y(y + 10) - 5(y + 10) = 0 \)
\( (y - 5)(y + 10) = 0 \)
This gives \( y = 5 \) or \( y = -10 \).
If \( y = 5 \), then \( x = 5 + 5 = 10 \).
If \( y = -10 \), then \( x = -10 + 5 = -5 \).
Therefore, the pairs of numbers are \( 10, 5 \) or \( -5, -10 \).
In simple words: Setting the difference to \( 5 \) and the product to \( 50 \) leads to a quadratic equation with solutions \( (10, 5) \) or \( (-5, -10) \).

Exam Tip: Since the smaller number has the larger reciprocal, the reciprocal equation is always set up as \( \frac{1}{\text{smaller}} - \frac{1}{\text{larger}} \).